I know that the voltage of a cell is calculated via the Nernst Equation as $$E_\text{cell}=E^\circ-\frac{RT}{nF}\ln Q_\mathrm r$$ where $Q_\mathrm r$ represents $\frac{\text{concentration at the anode}}{\text{concentration at the cathode}}$.

However I don't understand how these concentrations actually affect the potential difference. Could anybody explain the physical processes that mean changing the concentrations affect the voltage of the cell?


Potential is work to move charge and this work is done by the chemical energy stored in the reactants, so if the concentration of reactants increases then there are more reactants which will do more work so this implies that potential increases with comcentration.

  • $\begingroup$ Yes but when the concentration at the anode decreases the total cell voltage increases. To me this is counterintuitive and this is would I wanted to understand. $\endgroup$
    – Alex Jones
    Oct 9 '16 at 16:10
  • $\begingroup$ @AlexJones the concentration at anode which in the formula is probably of oxidised product which will not contribute to potential constructively and further helps in decreasing the potential. $\endgroup$
    – JM97
    Oct 10 '16 at 1:47

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